OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Exercise 7 (C)

S Chand Class 9 ICSE Maths Solutions Chapter 7 Logarithms Ex 7(C)

If \( \log 2 = 0.3010 \), \( \log 3 = 0.4771 \), \( \log 5 = 0.6990 \), \( \log 7 = 0.8451 \) and \( \log 11 \) are given, find the value of the following:

 

Question 1. log 6
Answer: We can write \( \log 6 \) as \( \log (2 \times 3) \). Using the logarithm product rule \( (\log(A \times B) = \log A + \log B) \), this becomes \( \log 2 + \log 3 \). Now, substitute the given values: \( 0.3010 + 0.4771 = 0.7781 \). So, \( \log 6 \) is \( 0.7781 \). Knowing the basic log values helps quickly solve many problems.
In simple words: To find \( \log 6 \), break it into \( \log 2 \) plus \( \log 3 \), then add their given values together.

🎯 Exam Tip: Remember that \( \log(A \times B) = \log A + \log B \). This is a fundamental property of logarithms used to break down expressions.

 

Question 2. log 12
Answer: First, express \( 12 \) as a product of its prime factors: \( 12 = 2 \times 2 \times 3 = 2^2 \times 3 \). So, \( \log 12 \) becomes \( \log (2^2 \times 3) \). Using the product rule, this is \( \log 2^2 + \log 3 \). Then, apply the power rule \( (\log A^n = n \log A) \) to get \( 2 \log 2 + \log 3 \). Substitute the given values: \( 2(0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791 \). Thus, \( \log 12 \) is \( 1.0791 \). Breaking numbers into prime factors simplifies logarithmic calculations.
In simple words: Change \( \log 12 \) to \( \log (2^2 \times 3) \). Then, use the rules to write it as \( 2 \log 2 + \log 3 \) and add the numbers.

🎯 Exam Tip: Always factorize the number inside the logarithm into prime factors first. This makes it easier to apply the product and power rules of logarithms.

 

Question 3. log 15
Answer: We can write \( \log 15 \) as \( \log (3 \times 5) \). Using the logarithm product rule, this becomes \( \log 3 + \log 5 \). Now, substitute the given values: \( 0.4771 + 0.6990 = 1.1761 \). Therefore, \( \log 15 \) is \( 1.1761 \). This method is very useful for numbers that are products of given log values.
In simple words: To find \( \log 15 \), write it as \( \log 3 \) plus \( \log 5 \), then add their given values.

🎯 Exam Tip: Simplify the number inside the logarithm by finding its prime factors to use the sum/difference properties of logarithms effectively.

 

Question 4. log 200
Answer: First, find the prime factorization of \( 200 \). We can divide \( 200 \) by \( 2 \) repeatedly until we get \( 2^3 \times 5^2 \). So, \( \log 200 \) becomes \( \log (2^3 \times 5^2) \). Using the product rule, this is \( \log 2^3 + \log 5^2 \). Applying the power rule gives \( 3 \log 2 + 2 \log 5 \). Substitute the given values: \( 3(0.3010) + 2(0.6990) = 0.9030 + 1.3980 = 2.3010 \). Thus, \( \log 200 \) is \( 2.3010 \). This process works for any composite number.
In simple words: Break \( 200 \) into its prime factors, which are \( 2^3 \times 5^2 \). Then, use the log rules to calculate \( 3 \log 2 + 2 \log 5 \) using the given values.

🎯 Exam Tip: When dealing with numbers like 100, 200, 1000, remember that they often involve powers of 10, 2, and 5, which are usually provided in logarithm problems.

 

Question 5. log 36
Answer: First, find the prime factorization of \( 36 \). We know \( 36 = 4 \times 9 = 2^2 \times 3^2 \). So, \( \log 36 \) becomes \( \log (2^2 \times 3^2) \). Using the product rule, this is \( \log 2^2 + \log 3^2 \). Applying the power rule gives \( 2 \log 2 + 2 \log 3 \). Substitute the given values: \( 2(0.3010) + 2(0.4771) = 0.6020 + 0.9542 = 1.5562 \). Therefore, \( \log 36 \) is \( 1.5562 \). Factorizing into squares often simplifies the problem significantly.
In simple words: Write \( 36 \) as \( 2^2 \times 3^2 \). Then, use log rules to get \( 2 \log 2 + 2 \log 3 \) and add the results.

🎯 Exam Tip: Always look for perfect squares or cubes when factorizing numbers to apply the power rule \( \log A^n = n \log A \) efficiently.

 

Question 6. log 80
Answer: First, find the prime factorization of \( 80 \). We know \( 80 = 16 \times 5 = 2^4 \times 5 \). So, \( \log 80 \) becomes \( \log (2^4 \times 5) \). Using the product rule, this is \( \log 2^4 + \log 5 \). Applying the power rule gives \( 4 \log 2 + \log 5 \). Substitute the given values: \( 4(0.3010) + 0.6990 = 1.2040 + 0.6990 = 1.9030 \). Thus, \( \log 80 \) is \( 1.9030 \). Prime factorization is a key step in these types of problems.
In simple words: Change \( \log 80 \) to \( \log (2^4 \times 5) \). Then, use log rules to make it \( 4 \log 2 + \log 5 \) and add the given values.

🎯 Exam Tip: If the number can be factored into a power of 2 and a prime, it often simplifies quickly using the product and power rules.

 

Question 7. log \( 2\frac { 1 }{ 3 } \)
Answer: First, convert the mixed fraction \( 2\frac { 1 }{ 3 } \) into an improper fraction: \( 2\frac { 1 }{ 3 } = \frac{(2 \times 3) + 1}{3} = \frac { 7 }{ 3 } \). So, the expression becomes \( \log \left(\frac{7}{3}\right) \). Using the logarithm quotient rule \( (\log(A/B) = \log A - \log B) \), this is \( \log 7 - \log 3 \). Substitute the given values: \( 0.8451 - 0.4771 = 0.3680 \). Therefore, \( \log 2\frac { 1 }{ 3 } \) is \( 0.3680 \). Converting mixed fractions to improper fractions is an important first step.
In simple words: First, change the mixed number \( 2\frac { 1 }{ 3 } \) to \( \frac { 7 }{ 3 } \). Then, use the rule that \( \log \frac{A}{B} \) is \( \log A - \log B \) to subtract \( \log 3 \) from \( \log 7 \).

🎯 Exam Tip: Always convert mixed fractions to improper fractions before applying logarithm rules to avoid errors in calculation.

 

Question 8. log \( 11^3 \)
Answer: Using the logarithm power rule \( (\log A^n = n \log A) \), \( \log 11^3 \) becomes \( 3 \log 11 \). The value of \( \log 11 \) is usually provided in such questions if it's not a standard base value. If it is provided, just multiply it by 3. This rule makes it much simpler to handle powers inside logarithms.
In simple words: When you have \( \log \) of a number with a power, just move the power to the front and multiply it by \( \log \) of the number. So, \( 3 \) times \( \log 11 \).

🎯 Exam Tip: The power rule \( \log A^n = n \log A \) is very useful for simplifying logarithmic expressions involving exponents.

 

Question 9. log \( \left(2\frac { 1 }{ 3 }\right)^5 \)
Answer: First, convert the mixed fraction \( 2\frac { 1 }{ 3 } \) to an improper fraction: \( \frac { 7 }{ 3 } \). So the expression is \( \log \left(\frac{7}{3}\right)^5 \). Using the power rule, this becomes \( 5 \log \left(\frac{7}{3}\right) \). Now, apply the quotient rule: \( 5[\log 7 - \log 3] \). Substitute the given values: \( 5[0.8451 - 0.4771] = 5(0.3680) = 1.8400 \). So, \( \log \left(2\frac { 1 }{ 3 }\right)^5 \) is \( 1.8400 \). Combining the power and quotient rules is a common approach.
In simple words: Change \( 2\frac { 1 }{ 3 } \) to \( \frac { 7 }{ 3 } \). Then move the power \( 5 \) to the front. Inside the bracket, write \( \log 7 - \log 3 \) and calculate, finally multiplying by \( 5 \).

🎯 Exam Tip: When multiple logarithm rules can be applied, it's often easiest to simplify fractions or powers first, then address sums or differences.

 

Question 10. If log 6 = 0.7782, find the value of log 36.
Answer: We are given \( \log 6 = 0.7782 \). We need to find \( \log 36 \). Notice that \( 36 \) can be written as \( 6^2 \). So, \( \log 36 = \log (6^2) \). Using the logarithm power rule, \( \log (6^2) \) becomes \( 2 \log 6 \). Now, substitute the given value of \( \log 6 \): \( 2(0.7782) = 1.5564 \). Thus, \( \log 36 \) is \( 1.5564 \). Recognizing perfect squares of numbers with known logs simplifies the calculation.
In simple words: Since \( 36 \) is \( 6^2 \), you can find \( \log 36 \) by just multiplying \( 2 \) with the given value of \( \log 6 \).

🎯 Exam Tip: Always check if the number in the new logarithm is a power of a number for which the logarithm is already given. This saves time and simplifies calculations.

 

Question 11. Given \( \log_{10} 25 = x \), \( \log_{10} 75 = y \), evaluate without using logarithmic tables, in terms of x and y.
(i) \( \log_{10} 3 \)
(ii) \( \log_{10} 2 \)
Answer:
(i) We have \( \log_{10} 25 = x \) and \( \log_{10} 75 = y \). To find \( \log_{10} 3 \), we can use the relationship \( 75 = 3 \times 25 \). So, \( \log_{10} 75 = \log_{10} (3 \times 25) \). Using the product rule, \( \log_{10} 75 = \log_{10} 3 + \log_{10} 25 \). Substitute the given values: \( y = \log_{10} 3 + x \). Rearranging, we get \( \log_{10} 3 = y - x \). This shows how to find a log value from related expressions.
(ii) We know \( \log_{10} 25 = x \). Also, \( 25 = 5^2 \), so \( \log_{10} 5^2 = x \). Using the power rule, \( 2 \log_{10} 5 = x \), which means \( \log_{10} 5 = \frac{x}{2} \). Now, to find \( \log_{10} 2 \), we can use the fact that \( \log_{10} 10 = 1 \). We know \( 10 = 2 \times 5 \), so \( \log_{10} 10 = \log_{10} (2 \times 5) = \log_{10} 2 + \log_{10} 5 \). Substitute \( 1 = \log_{10} 2 + \frac{x}{2} \). Rearranging, \( \log_{10} 2 = 1 - \frac{x}{2} = \frac{2-x}{2} \). It is always useful to know \( \log_{10} 2 \) or \( \log_{10} 5 \) when \( \log_{10} 10 \) is involved.
In simple words:
(i) To find \( \log_{10} 3 \), notice that \( 75 \) is \( 3 \) times \( 25 \). So, \( \log_{10} 3 \) is \( \log_{10} 75 - \log_{10} 25 \), which is \( y - x \).
(ii) For \( \log_{10} 2 \), first find \( \log_{10} 5 \) from \( \log_{10} 25 = x \). Then use \( \log_{10} 10 = \log_{10} 2 + \log_{10} 5 \) to find \( \log_{10} 2 \).

🎯 Exam Tip: Remember the basic logarithmic identities: \( \log_{b} (MN) = \log_{b} M + \log_{b} N \), \( \log_{b} (M/N) = \log_{b} M - \log_{b} N \), and \( \log_{b} M^k = k \log_{b} M \). Also, \( \log_{10} 10 = 1 \).

 

Question 12. Given \( \log 31.87 = x \), write down in terms of x.
(i) \( \log (31.87)^2 \)
(ii) \( \log_{10} 0.03187 \)
(iii) \( \log_{10} \sqrt{31870} \)
Answer: We are given \( \log 31.87 = x \).
(i) Using the power rule \( (\log A^n = n \log A) \), \( \log (31.87)^2 = 2 \log (31.87) \). Since \( \log 31.87 = x \), this becomes \( 2x \). This shows how powers multiply the logarithm.
(ii) To find \( \log_{10} 0.03187 \), we can write \( 0.03187 = \frac{31.87}{1000} \). Using the quotient rule \( (\log(A/B) = \log A - \log B) \), this is \( \log_{10} 31.87 - \log_{10} 1000 \). We know \( \log_{10} 31.87 = x \) and \( \log_{10} 1000 = 3 \) (because \( 10^3 = 1000 \)). So, the expression becomes \( x - 3 \). This is a common way to handle decimals.
(iii) To find \( \log_{10} \sqrt{31870} \), first write the square root as a power: \( \sqrt{31870} = (31870)^{1/2} \). Also, \( 31870 = 31.87 \times 1000 \). So, the expression is \( \log_{10} (31.87 \times 1000)^{1/2} \). Using the power rule, this is \( \frac{1}{2} \log_{10} (31.87 \times 1000) \). Now, apply the product rule: \( \frac{1}{2} [\log_{10} 31.87 + \log_{10} 1000] \). Substitute the given values: \( \frac{1}{2} [x + 3] = \frac{x+3}{2} \). Combining these rules is often necessary.
In simple words:
(i) For \( \log (31.87)^2 \), just multiply \( x \) by \( 2 \).
(ii) For \( \log_{10} 0.03187 \), think of it as \( \log_{10} (31.87 \div 1000) \). This becomes \( \log_{10} 31.87 - \log_{10} 1000 \), which is \( x - 3 \).
(iii) For \( \log_{10} \sqrt{31870} \), rewrite \( \sqrt{31870} \) as \( (31.87 \times 1000)^{1/2} \). Use the power and product rules to get \( \frac{1}{2} (x + 3) \).

🎯 Exam Tip: Remember that \( \log_{10} 10^n = n \) and \( \sqrt{A} = A^{1/2} \). These help simplify expressions involving powers of 10 and square roots.

 

Question 13. Solve the equation
(i) \( \log_{10} (x + 1) + \log_{10} (x – 1) = \log_{10} 11 + 2 \log_{10} 3 \)
(ii) \( \log (10x + 5) – \log (x – 4) = 2 \)
Answer:
(i) Given \( \log_{10} (x + 1) + \log_{10} (x – 1) = \log_{10} 11 + 2 \log_{10} 3 \).
On the left side, use the product rule: \( \log_{10} [(x + 1)(x – 1)] \). This simplifies to \( \log_{10} (x^2 – 1) \).
On the right side, use the power rule for \( 2 \log_{10} 3 \) to get \( \log_{10} 3^2 = \log_{10} 9 \). So, the right side is \( \log_{10} 11 + \log_{10} 9 \). Using the product rule, this is \( \log_{10} (11 \times 9) = \log_{10} 99 \).
Now the equation is \( \log_{10} (x^2 – 1) = \log_{10} 99 \).
Since the logarithms are equal and have the same base, their arguments must be equal: \( x^2 – 1 = 99 \).
Add \( 1 \) to both sides: \( x^2 = 99 + 1 = 100 \).
Take the square root of both sides: \( x = \sqrt{100} = 10 \). Remember that \( x \) must be a positive value in logarithmic expressions, and \( (x+1) \) and \( (x-1) \) must be positive. If \( x=10 \), then \( x+1 = 11 \) and \( x-1=9 \), both positive. So \( x=10 \) is the correct solution.
In simple words:
(i) Combine the left side as \( \log_{10} (x^2 - 1) \). Combine the right side as \( \log_{10} (11 \times 3^2) \). Then, set \( x^2 - 1 \) equal to \( 99 \) and solve for \( x \).

🎯 Exam Tip: Always check if the values of x make the arguments of the logarithms positive. For \( \log_b A \), \( A \) must be greater than \( 0 \).

 

Question 13. Solve the equation
(ii) \( \log (10x + 5) – \log (x – 4) = 2 \)
Answer:
(ii) Given \( \log (10x + 5) – \log (x – 4) = 2 \).
On the left side, use the quotient rule: \( \log \left(\frac{10x+5}{x-4}\right) \).
On the right side, express \( 2 \) as a logarithm to base 10 (since no base is specified, base 10 is implied): \( 2 = \log_{10} 100 \).
Now the equation is \( \log \left(\frac{10x+5}{x-4}\right) = \log_{10} 100 \).
Equate the arguments: \( \frac{10x+5}{x-4} = 100 \).
Multiply both sides by \( (x – 4) \): \( 10x + 5 = 100(x – 4) \).
Distribute \( 100 \): \( 10x + 5 = 100x – 400 \).
Subtract \( 10x \) from both sides: \( 5 = 90x – 400 \).
Add \( 400 \) to both sides: \( 405 = 90x \).
Divide by \( 90 \): \( x = \frac{405}{90} = \frac{45}{10} = 4.5 \). When solving equations, remember to ensure the base of the logarithm is consistent.
In simple words:
(ii) Combine the left side as \( \log \frac{10x+5}{x-4} \). Change \( 2 \) to \( \log 100 \). Set the fractions equal and solve for \( x \).

🎯 Exam Tip: When an equation has a constant on one side, convert it to a logarithm with the same base as the other terms (e.g., \( 2 = \log_{10} 100 \)).

 

Question 14.
(a) Given \( 2 \log_{10} x + \frac { 1 }{ 2 } \log_{10} y = 1 \), express y in terms of x.
(b) Express as a single logarithm : \( 2 \log 3 - \frac { 1 }{ 2 } \log 16 + \log 12 \)
Answer:
(a) Given \( 2 \log_{10} x + \frac { 1 }{ 2 } \log_{10} y = 1 \).
Use the power rule: \( \log_{10} x^2 + \log_{10} y^{1/2} = 1 \).
Use the product rule: \( \log_{10} (x^2 y^{1/2}) = 1 \).
Convert the logarithm to exponential form: \( x^2 y^{1/2} = 10^1 \).
So, \( x^2 \sqrt{y} = 10 \).
To express \( y \) in terms of \( x \), isolate \( \sqrt{y} \): \( \sqrt{y} = \frac{10}{x^2} \).
Square both sides: \( y = \left(\frac{10}{x^2}\right)^2 = \frac{100}{x^4} \). This process converts the logarithmic equation into an algebraic one.
(b) Given \( 2 \log 3 - \frac { 1 }{ 2 } \log 16 + \log 12 \).
First, apply the power rule: \( \log 3^2 - \log 16^{1/2} + \log 12 \).
This simplifies to \( \log 9 - \log 4 + \log 12 \).
Now, use the quotient rule for the first two terms: \( \log \left(\frac{9}{4}\right) + \log 12 \).
Finally, use the product rule: \( \log \left(\frac{9}{4} \times 12\right) \).
Simplify the expression inside the logarithm: \( \frac{9}{4} \times 12 = 9 \times 3 = 27 \).
So, the expression as a single logarithm is \( \log 27 \). Combining terms requires careful application of the rules.
In simple words:
(a) Move the powers to the logs, then combine them into a single \( \log_{10} \) term. Convert to an exponent form, then solve for \( y \).
(b) Move the numbers in front of logs to become powers. Then, use the rules \( \log A - \log B = \log \frac{A}{B} \) and \( \log A + \log B = \log (A \times B) \) to combine all terms into one \( \log \).

🎯 Exam Tip: Remember that \( \log_{10} A = B \) is the same as \( A = 10^B \). This conversion is crucial when solving for variables within logarithmic equations.

 

Question 15. Given that \( \log_{10} y + 2 \log_{10} x = 2 \), express y in terms of x.
Answer: Given \( \log_{10} y + 2 \log_{10} x = 2 \).
Apply the power rule to the second term: \( \log_{10} y + \log_{10} x^2 = 2 \).
Apply the product rule to combine the terms on the left: \( \log_{10} (y \times x^2) = 2 \).
Now, convert the logarithmic equation into exponential form. Remember that \( \log_b A = C \) means \( b^C = A \). Here, \( b=10 \), \( C=2 \), and \( A = yx^2 \).
So, \( yx^2 = 10^2 \).
This gives \( yx^2 = 100 \).
To express \( y \) in terms of \( x \), divide both sides by \( x^2 \): \( y = \frac{100}{x^2} \). This is a common method for rearranging logarithmic equations.
In simple words: First, rewrite \( 2 \log_{10} x \) as \( \log_{10} x^2 \). Then combine the logs on the left side to get \( \log_{10} (yx^2) \). Change the equation into exponential form: \( yx^2 = 10^2 \). Finally, solve for \( y \) by dividing by \( x^2 \).

🎯 Exam Tip: Always make sure to isolate the variable you want to express in terms of others, using the inverse operations of logarithms and exponents.

 

Question 16. If \( a = 1 + \log_{10} 2 – \log_{10} 5 \), \( b = 2 \log_{10} 3 \) and \( c = \log_{10} m – \log_{10} 5 \), find the value of m if \( a + b = 2c \) (Do not use log tables).
Answer:
First, simplify \( a \):
\( a = 1 + \log_{10} 2 – \log_{10} 5 \)
Since \( 1 = \log_{10} 10 \), substitute this: \( a = \log_{10} 10 + \log_{10} 2 – \log_{10} 5 \).
Using the product rule and quotient rule: \( a = \log_{10} \left(\frac{10 \times 2}{5}\right) = \log_{10} \left(\frac{20}{5}\right) = \log_{10} 4 \).

Next, simplify \( b \):
\( b = 2 \log_{10} 3 \).
Using the power rule: \( b = \log_{10} 3^2 = \log_{10} 9 \).

Next, simplify \( c \):
\( c = \log_{10} m – \log_{10} 5 \).
Using the quotient rule: \( c = \log_{10} \left(\frac{m}{5}\right) \).

Now, use the condition \( a + b = 2c \):
\( \log_{10} 4 + \log_{10} 9 = 2 \log_{10} \left(\frac{m}{5}\right) \).
On the left side, use the product rule: \( \log_{10} (4 \times 9) = \log_{10} 36 \).
On the right side, use the power rule: \( 2 \log_{10} \left(\frac{m}{5}\right) = \log_{10} \left(\frac{m}{5}\right)^2 = \log_{10} \left(\frac{m^2}{25}\right) \).
So, the equation becomes \( \log_{10} 36 = \log_{10} \left(\frac{m^2}{25}\right) \).
Since the logarithms are equal and have the same base, their arguments must be equal:
\( 36 = \frac{m^2}{25} \).
Multiply both sides by \( 25 \): \( m^2 = 36 \times 25 = 900 \).
Take the square root of both sides: \( m = \sqrt{900} = 30 \). Since \( m \) is an argument of a logarithm, it must be positive. This example shows solving for a variable inside a logarithm by simplifying expressions first.
In simple words: First, simplify \( a \), \( b \), and \( c \) into single \( \log_{10} \) terms. Then, put them into the equation \( a + b = 2c \) and use log rules to simplify both sides. After that, set the numbers inside the \( \log \) equal to each other and solve for \( m \).

🎯 Exam Tip: When given an equation with multiple logarithmic terms, simplify each term into a single logarithm first before substituting them into the main equation. Remember that \( \log_{10} 10 = 1 \).

 

Question 17. Express as a single logarithm: \( 2 + \frac { 1 }{ 2 } \log_{10} 9 – 2 \log_{10} 5 \)
Answer: Given expression: \( 2 + \frac { 1 }{ 2 } \log_{10} 9 – 2 \log_{10} 5 \).
First, convert the constant \( 2 \) to a logarithm with base 10: \( 2 = \log_{10} 10^2 = \log_{10} 100 \).
Apply the power rule to the other terms: \( \frac { 1 }{ 2 } \log_{10} 9 = \log_{10} 9^{1/2} = \log_{10} 3 \) and \( 2 \log_{10} 5 = \log_{10} 5^2 = \log_{10} 25 \).
Substitute these back into the expression: \( \log_{10} 100 + \log_{10} 3 - \log_{10} 25 \).
Now, use the product rule for the first two terms: \( \log_{10} (100 \times 3) - \log_{10} 25 = \log_{10} 300 - \log_{10} 25 \).
Finally, use the quotient rule: \( \log_{10} \left(\frac{300}{25}\right) \).
Simplify the fraction: \( \frac{300}{25} = 12 \).
So, the expression as a single logarithm is \( \log_{10} 12 \). Converting constants to logarithms and then applying rules simplifies the expression.
In simple words: Change \( 2 \) to \( \log_{10} 100 \). Move the numbers in front of the other logs to become powers. Then, use the rules for adding and subtracting logs to combine everything into one \( \log_{10} \) term.

🎯 Exam Tip: Always convert any constant terms into logarithms of the appropriate base (e.g., \( C = \log_b b^C \)) before combining with other logarithmic terms.

 

Question 18. If \( a = \log 12 \), \( b = \log 6 \) and \( c = 2 \log \sqrt{2} \), find
(i) \( a-b-c \)
(ii) \( 9^{a-b-c} \)
Answer:
First, simplify \( c \):
\( c = 2 \log \sqrt{2} = 2 \log 2^{1/2} \).
Using the power rule: \( c = 2 \times \frac{1}{2} \log 2 = \log 2 \).

(i) Now, find \( a - b - c \):
\( a - b - c = \log 12 - \log 6 - \log 2 \).
Use the quotient rule repeatedly: \( a - b - c = \log \left(\frac{12}{6 \times 2}\right) \).
Simplify the expression inside the logarithm: \( \frac{12}{12} = 1 \).
So, \( a - b - c = \log 1 \).
We know that \( \log 1 = 0 \) for any base. Therefore, \( a - b - c = 0 \). This is a fundamental property of logarithms.

(ii) Now, find \( 9^{a-b-c} \):
From part (i), we found that \( a - b - c = 0 \).
So, \( 9^{a-b-c} = 9^0 \).
Any non-zero number raised to the power of \( 0 \) is \( 1 \). Therefore, \( 9^0 = 1 \). This uses basic exponent rules.
In simple words:
First, simplify \( c \) to \( \log 2 \).
(i) For \( a-b-c \), write it as \( \log 12 - \log 6 - \log 2 \). Use the rule for subtracting logs to get \( \log \left(\frac{12}{6 \times 2}\right) \). This simplifies to \( \log 1 \), which is \( 0 \).
(ii) Since \( a-b-c \) is \( 0 \), \( 9^{a-b-c} \) means \( 9^0 \), which is \( 1 \).

🎯 Exam Tip: Remember two key properties: \( \log_b 1 = 0 \) for any base \( b \neq 1 \), and any non-zero number raised to the power of zero is \( 1 \).

 

Question 19. If \( x = \log_{10} 12 \), \( y = \log_4 2 \times \log_{10} 9 \) and \( z = \log_{10} (0.4) \) then find
(i) \( x - y - z \)
Answer:
First, simplify \( x \):
\( x = \log_{10} 12 = \log_{10} (2^2 \times 3) = \log_{10} 2^2 + \log_{10} 3 = 2 \log_{10} 2 + \log_{10} 3 \).

Next, simplify \( y \):
\( y = \log_4 2 \times \log_{10} 9 \).
For \( \log_4 2 \): We know \( 4 = 2^2 \), so \( \log_4 2 = \log_{2^2} 2 \). If \( \log_{b^n} A = \frac{1}{n} \log_b A \), then \( \log_{2^2} 2 = \frac{1}{2} \log_2 2 = \frac{1}{2} \times 1 = \frac{1}{2} \). Alternatively, if \( \log_4 2 = k \), then \( 4^k = 2 \implies (2^2)^k = 2^1 \implies 2k = 1 \implies k = \frac{1}{2} \).
For \( \log_{10} 9 \): \( \log_{10} 9 = \log_{10} 3^2 = 2 \log_{10} 3 \).
So, \( y = \frac{1}{2} \times 2 \log_{10} 3 = \log_{10} 3 \).

Next, simplify \( z \):
\( z = \log_{10} (0.4) = \log_{10} \left(\frac{4}{10}\right) = \log_{10} 4 - \log_{10} 10 \).
We know \( \log_{10} 4 = \log_{10} 2^2 = 2 \log_{10} 2 \), and \( \log_{10} 10 = 1 \).
So, \( z = 2 \log_{10} 2 - 1 \).

(i) Now, find \( x - y - z \):
\( x - y - z = (2 \log_{10} 2 + \log_{10} 3) - (\log_{10} 3) - (2 \log_{10} 2 - 1) \).
Distribute the negative sign: \( 2 \log_{10} 2 + \log_{10} 3 - \log_{10} 3 - 2 \log_{10} 2 + 1 \).
Combine like terms: \( (2 \log_{10} 2 - 2 \log_{10} 2) + (\log_{10} 3 - \log_{10} 3) + 1 \).
All the logarithmic terms cancel out, leaving: \( 0 + 0 + 1 = 1 \).
Thus, \( x - y - z = 1 \). Simplifying each variable first makes the final calculation easier.
In simple words: First, break down \( x \), \( y \), and \( z \) into their simplest log forms. Then, carefully subtract them, making sure to cancel out terms that are the same but opposite. The final answer is \( 1 \).

🎯 Exam Tip: When dealing with multiple log expressions, simplify each term (like x, y, z) individually before combining them. Pay close attention to base changes (like \( \log_4 2 \)) and decimal conversions.

 

Question 19. (cont.)
(ii) \( 6^{x-y-z} \)
Answer:
(ii) From part (i), we proved that \( x - y - z = 1 \).
So, \( 6^{x-y-z} \) becomes \( 6^1 \).
Any number raised to the power of \( 1 \) is the number itself. Therefore, \( 6^1 = 6 \). This demonstrates the application of basic exponent properties after simplifying the logarithmic expression.
In simple words: Since we found \( x-y-z \) is \( 1 \), \( 6^{x-y-z} \) just means \( 6^1 \), which is \( 6 \).

🎯 Exam Tip: Always relate previous parts of a question to subsequent parts, as they often build upon earlier results to simplify calculations.

 

Question 20. If \( p = \log_{10} 20 \) and \( q = \log_{10} 25 \), find x such that \( 2 \log_{10} (x + 1) = 2p - q \).
Answer:
First, simplify \( p \) and \( q \):
\( p = \log_{10} 20 = \log_{10} (2^2 \times 5) = 2 \log_{10} 2 + \log_{10} 5 \).
\( q = \log_{10} 25 = \log_{10} 5^2 = 2 \log_{10} 5 \).

Next, simplify the right side of the main equation, \( 2p - q \):
\( 2p - q = 2(2 \log_{10} 2 + \log_{10} 5) - (2 \log_{10} 5) \).
Distribute the \( 2 \): \( 4 \log_{10} 2 + 2 \log_{10} 5 - 2 \log_{10} 5 \).
The terms \( 2 \log_{10} 5 \) and \( -2 \log_{10} 5 \) cancel out, leaving \( 4 \log_{10} 2 \).
Using the power rule: \( 4 \log_{10} 2 = \log_{10} 2^4 = \log_{10} 16 \).

Now, substitute this back into the main equation: \( 2 \log_{10} (x + 1) = \log_{10} 16 \).
Apply the power rule to the left side: \( \log_{10} (x + 1)^2 = \log_{10} 16 \).
Since the logarithms are equal and have the same base, their arguments must be equal:
\( (x + 1)^2 = 16 \).
Take the square root of both sides: \( x + 1 = \pm \sqrt{16} \).
So, \( x + 1 = 4 \) or \( x + 1 = -4 \).
If \( x + 1 = 4 \), then \( x = 3 \).
If \( x + 1 = -4 \), then \( x = -5 \).
However, the argument of a logarithm must be positive. If \( x = -5 \), then \( x + 1 = -4 \), which is not allowed. So, \( x = -5 \) is an extraneous solution.
Therefore, \( x = 3 \) is the only valid solution. Always check solutions against the domain of the logarithm.
In simple words: First, simplify \( p \) and \( q \). Then, calculate \( 2p - q \) using log rules, which simplifies to \( \log_{10} 16 \). Set \( 2 \log_{10} (x+1) \) equal to \( \log_{10} 16 \), simplify both sides, and solve for \( x \). Remember to discard any \( x \) values that make the log argument negative.

🎯 Exam Tip: After solving for \( x \), always verify that the value(s) obtained do not make the argument of any logarithm in the original equation negative or zero.

 

Question 21. Without using logarithm tables, evaluate : \( 3 + \log_{10} (10^{-2}) \)
Answer: Given expression: \( 3 + \log_{10} (10^{-2}) \).
Use the power rule for logarithms: \( \log_{10} (10^{-2}) = -2 \log_{10} 10 \).
We know that \( \log_{10} 10 = 1 \).
So, \( \log_{10} (10^{-2}) = -2 \times 1 = -2 \).
Substitute this back into the expression: \( 3 + (-2) \).
Calculate the final value: \( 3 - 2 = 1 \). This demonstrates how to simplify expressions using basic log properties.
In simple words: First, simplify \( \log_{10} (10^{-2}) \) to \( -2 \) because \( \log_{10} 10 \) is \( 1 \). Then, add \( 3 \) and \( -2 \) to get \( 1 \).

🎯 Exam Tip: Remember the identity \( \log_b b^x = x \). This property allows for direct evaluation of terms like \( \log_{10} 10^{-2} \).

 

Question 22. Given \( \log_{10} x = a \), \( \log_{10} y = b \)
(i) Write down \( 10^{a-1} \) in terms of x
(ii) Write down \( 10^{2b} \) in terms of y
(iii) If \( \log_{10} P = 2a – b \), express P in terms of x and y.
Answer: We are given \( \log_{10} x = a \) and \( \log_{10} y = b \). From these, we can write the exponential forms: \( 10^a = x \) and \( 10^b = y \).
(i) To express \( 10^{a-1} \) in terms of \( x \):
Using exponent rules, \( 10^{a-1} = \frac{10^a}{10^1} \).
Substitute \( 10^a = x \): \( 10^{a-1} = \frac{x}{10} \). This simplifies the expression by replacing the exponential term.
(ii) To express \( 10^{2b} \) in terms of \( y \):
Using exponent rules, \( 10^{2b} = (10^b)^2 \).
Substitute \( 10^b = y \): \( 10^{2b} = y^2 \). This shows how to handle powers of exponential terms.
(iii) Given \( \log_{10} P = 2a – b \). To express \( P \) in terms of \( x \) and \( y \):
Substitute \( a = \log_{10} x \) and \( b = \log_{10} y \) into the equation:
\( \log_{10} P = 2 \log_{10} x - \log_{10} y \).
Apply the power rule: \( \log_{10} P = \log_{10} x^2 - \log_{10} y \).
Apply the quotient rule: \( \log_{10} P = \log_{10} \left(\frac{x^2}{y}\right) \).
Since the logarithms are equal and have the same base, their arguments must be equal:
\( P = \frac{x^2}{y} \). This process links logarithmic and exponential forms.
In simple words: First, convert \( \log_{10} x = a \) to \( 10^a = x \) and \( \log_{10} y = b \) to \( 10^b = y \).
(i) For \( 10^{a-1} \), use the rule \( A^{B-C} = A^B \div A^C \) to get \( \frac{10^a}{10^1} \), then replace \( 10^a \) with \( x \).
(ii) For \( 10^{2b} \), write it as \( (10^b)^2 \), then replace \( 10^b \) with \( y \).
(iii) For \( \log_{10} P = 2a – b \), replace \( a \) and \( b \) with their log forms. Then simplify using log rules to find \( P \).

🎯 Exam Tip: Always remember the definition of logarithms in terms of exponents: \( \log_b A = C \iff b^C = A \). This helps in converting between the two forms easily.

 

Question 23. Simplify without using tables : \( 2 \log_{10} 5 + \log_{10} 8 - \frac{1}{2} \log_{10} 4 \)
Answer: Given expression: \( 2 \log_{10} 5 + \log_{10} 8 - \frac{1}{2} \log_{10} 4 \).
Apply the power rule to the first and third terms:
\( 2 \log_{10} 5 = \log_{10} 5^2 = \log_{10} 25 \).
\( \frac{1}{2} \log_{10} 4 = \log_{10} 4^{1/2} = \log_{10} \sqrt{4} = \log_{10} 2 \).
Substitute these back into the expression:
\( \log_{10} 25 + \log_{10} 8 - \log_{10} 2 \).
Apply the product rule for the first two terms:
\( \log_{10} (25 \times 8) - \log_{10} 2 = \log_{10} 200 - \log_{10} 2 \).
Apply the quotient rule:
\( \log_{10} \left(\frac{200}{2}\right) = \log_{10} 100 \).
We know that \( \log_{10} 100 = \log_{10} 10^2 = 2 \).
So, the simplified value is \( 2 \). Combining all the log rules step-by-step is crucial for correct simplification.
In simple words: Move the numbers in front of logs to become powers. Then, use the rules for adding and subtracting logs to combine everything into one \( \log_{10} \) term. Finally, simplify the number inside the log to find the answer.

🎯 Exam Tip: Always simplify terms with fractional or integer coefficients using the power rule first, then combine using product and quotient rules. Remember \( \log_{10} 10^n = n \).

 

Question 24. Given that \( \log_{10} 2 = x \), \( \log_{10} 3 = y \), find
(i) \( \log_{10} 60 \)
(ii) \( \log_{10} 1.2 \) in terms of x and y
Answer: We are given \( \log_{10} 2 = x \) and \( \log_{10} 3 = y \).
(i) To find \( \log_{10} 60 \):
First, find the prime factorization of \( 60 \): \( 60 = 2 \times 3 \times 10 \).
So, \( \log_{10} 60 = \log_{10} (2 \times 3 \times 10) \).
Using the product rule: \( \log_{10} 2 + \log_{10} 3 + \log_{10} 10 \).
Substitute the given values and \( \log_{10} 10 = 1 \): \( x + y + 1 \).
Therefore, \( \log_{10} 60 = x + y + 1 \). Breaking numbers into primes and powers of 10 helps a lot.
(ii) To find \( \log_{10} 1.2 \) in terms of \( x \) and \( y \):
First, write \( 1.2 \) as a fraction: \( 1.2 = \frac{12}{10} \).
So, \( \log_{10} 1.2 = \log_{10} \left(\frac{12}{10}\right) \).
Using the quotient rule: \( \log_{10} 12 - \log_{10} 10 \).
Now, find \( \log_{10} 12 \): \( \log_{10} 12 = \log_{10} (2^2 \times 3) = \log_{10} 2^2 + \log_{10} 3 = 2 \log_{10} 2 + \log_{10} 3 \).
Substitute \( x \) and \( y \): \( \log_{10} 12 = 2x + y \).
And we know \( \log_{10} 10 = 1 \).
So, \( \log_{10} 1.2 = (2x + y) - 1 \). This involves converting decimals to fractions before applying log rules.
In simple words:
(i) For \( \log_{10} 60 \), write \( 60 \) as \( 2 \times 3 \times 10 \). Then, use the rule for adding logs and replace \( \log_{10} 2 \) with \( x \), \( \log_{10} 3 \) with \( y \), and \( \log_{10} 10 \) with \( 1 \).
(ii) For \( \log_{10} 1.2 \), change \( 1.2 \) to \( \frac{12}{10} \). Then use the rule for subtracting logs to get \( \log_{10} 12 - \log_{10} 10 \). Next, break \( \log_{10} 12 \) into \( 2x + y \), then subtract \( 1 \).

🎯 Exam Tip: For decimals, always convert them to simple fractions (e.g., \( 0.4 = 4/10 \), \( 1.2 = 12/10 \)) before applying the quotient rule of logarithms.

 

Question 25. Given \( 2 \log_{10} x + 1 = \log_{10} 250 \), find
(i) x
(ii) \( \log_{10} 2x \)
Answer: Given equation: \( 2 \log_{10} x + 1 = \log_{10} 250 \).
First, rewrite the equation by converting \( 1 \) to a logarithm with base 10: \( 1 = \log_{10} 10 \).
So, \( 2 \log_{10} x + \log_{10} 10 = \log_{10} 250 \).
Apply the power rule to the first term: \( \log_{10} x^2 + \log_{10} 10 = \log_{10} 250 \).
Apply the product rule to the left side: \( \log_{10} (x^2 \times 10) = \log_{10} 250 \).
So, \( \log_{10} (10x^2) = \log_{10} 250 \).
Since the logarithms are equal and have the same base, their arguments must be equal:
\( 10x^2 = 250 \).
Divide by \( 10 \): \( x^2 = 25 \).
Take the square root of both sides: \( x = \pm \sqrt{25} \).
So, \( x = 5 \) or \( x = -5 \).
However, the argument of a logarithm must be positive, so \( x \) must be greater than \( 0 \). Therefore, \( x = -5 \) is an extraneous solution. The valid solution is \( x = 5 \). This requires careful checking of valid domains.

(i) From the calculation above, \( x = 5 \).

(ii) Now, find \( \log_{10} 2x \):
Substitute the value of \( x = 5 \) into the expression: \( \log_{10} (2 \times 5) \).
This simplifies to \( \log_{10} 10 \).
We know that \( \log_{10} 10 = 1 \).
Therefore, \( \log_{10} 2x = 1 \). This is a direct substitution after solving for x.
In simple words:
First, change \( 1 \) to \( \log_{10} 10 \). Combine the left side into one \( \log_{10} \) term, then set it equal to \( \log_{10} 250 \). Solve for \( x \) and make sure \( x \) is positive.
(i) The value of \( x \) is \( 5 \).
(ii) Put \( x=5 \) into \( \log_{10} 2x \) to get \( \log_{10} 10 \), which equals \( 1 \).

🎯 Exam Tip: Always convert constant terms to logarithms of the same base as other terms in the equation to simplify them effectively. Also, remember to check for extraneous solutions.

 

Question 26.
(a) Given that \( \log x = m + n \) and \( \log y = m – n \), express the value of \( \log_{10} \frac{10x}{y^2} \) in terms of m and n.
(b) Solve for x : \( \log_{10} x = - 2 \).
Answer:
(a) Given \( \log x = m + n \) and \( \log y = m – n \).
We need to express \( \log_{10} \frac{10x}{y^2} \) in terms of \( m \) and \( n \).
Apply the quotient rule: \( \log_{10} \frac{10x}{y^2} = \log_{10} (10x) - \log_{10} y^2 \).
Apply the product rule to the first term and the power rule to the second term:
\( \log_{10} 10 + \log_{10} x - 2 \log_{10} y \).
Substitute \( \log_{10} 10 = 1 \), \( \log x = m + n \), and \( \log y = m – n \):
\( 1 + (m + n) - 2(m – n) \).
Distribute the \( -2 \): \( 1 + m + n - 2m + 2n \).
Combine like terms: \( 1 + (m - 2m) + (n + 2n) = 1 - m + 3n \).
So, \( \log_{10} \frac{10x}{y^2} = 1 - m + 3n \). This showcases combining multiple log rules.
(b) Given \( \log_{10} x = - 2 \).
To solve for \( x \), convert the logarithmic equation to its exponential form. Remember that \( \log_b A = C \) means \( b^C = A \). Here, \( b = 10 \), \( C = -2 \), and \( A = x \).
So, \( x = 10^{-2} \).
We know that \( 10^{-2} = \frac{1}{10^2} = \frac{1}{100} \).
Therefore, \( x = \frac{1}{100} \). This is a direct conversion from log to exponent form.
In simple words:
(a) Break down \( \log_{10} \frac{10x}{y^2} \) using the rules for division, multiplication, and powers of logs. Then, replace \( \log x \) with \( (m+n) \) and \( \log y \) with \( (m-n) \), and \( \log_{10} 10 \) with \( 1 \), then simplify the expression.
(b) For \( \log_{10} x = -2 \), change it to an exponent form: \( x = 10^{-2} \). This means \( x \) is \( \frac{1}{100} \).

🎯 Exam Tip: Be careful with signs when distributing negative numbers, especially in expressions like \( -2(m-n) \). Ensure to convert from logarithmic to exponential form correctly when solving for variables.

 

Question 27. If \( \log \left(\frac{p+q}{3}\right)=\frac{1}{2} (\log p + \log q) \) prove that \( p^2 + q^2 = 7pq \).
Answer: Given \( \log \left(\frac{p+q}{3}\right)=\frac{1}{2} (\log p + \log q) \).
On the right side, first apply the product rule: \( \log p + \log q = \log (pq) \).
Then, apply the power rule: \( \frac{1}{2} \log (pq) = \log (pq)^{1/2} = \log \sqrt{pq} \).
So, the equation becomes \( \log \left(\frac{p+q}{3}\right) = \log \sqrt{pq} \).
Since the logarithms are equal and have the same base, their arguments must be equal:
\( \frac{p+q}{3} = \sqrt{pq} \).
Now, square both sides to remove the square root:
\( \left(\frac{p+q}{3}\right)^2 = (\sqrt{pq})^2 \).
\( \frac{(p+q)^2}{3^2} = pq \).
\( \frac{p^2 + 2pq + q^2}{9} = pq \).
Multiply both sides by \( 9 \): \( p^2 + 2pq + q^2 = 9pq \).
Subtract \( 2pq \) from both sides: \( p^2 + q^2 = 9pq - 2pq \).
\( p^2 + q^2 = 7pq \).
This proves the required identity. The critical step is squaring both sides to eliminate the square root and simplifying.
In simple words: Start by using log rules to simplify the right side of the given equation to \( \log \sqrt{pq} \). Then, set the things inside the logs equal: \( \frac{p+q}{3} = \sqrt{pq} \). Next, square both sides to remove the square root and simplify the equation until you get \( p^2 + q^2 = 7pq \).

🎯 Exam Tip: When you have a square root in an equation, squaring both sides is often the next step to simplify and solve or prove an identity. Be careful to square the entire expression on both sides.

 

Question 28. If \( x^2 + y^2 = 51xy \), prove that \( \log \frac{x-y}{7} = \frac{1}{2} (\log x + \log y) \).
Answer: Given \( x^2 + y^2 = 51xy \). We need to prove \( \log \frac{x-y}{7} = \frac{1}{2} (\log x + \log y) \).
Let's start by manipulating the given equation to get a term like \( (x-y)^2 \).
Subtract \( 2xy \) from both sides: \( x^2 + y^2 - 2xy = 51xy - 2xy \).
This simplifies to \( (x - y)^2 = 49xy \).
Divide both sides by \( 49 \): \( \frac{(x - y)^2}{49} = xy \).
Rewrite \( 49 \) as \( 7^2 \): \( \frac{(x - y)^2}{7^2} = xy \).
This can be written as \( \left(\frac{x - y}{7}\right)^2 = xy \).
Now, take the square root of both sides. Since \( x \) and \( y \) are arguments of logarithms, they must be positive. Also, \( x-y \) must be positive for \( \log \frac{x-y}{7} \) to be defined. So, we take the positive square root:
\( \frac{x - y}{7} = \sqrt{xy} \).
Now, apply logarithm to both sides (assuming base 10 for \( \log \)):
\( \log \left(\frac{x - y}{7}\right) = \log (\sqrt{xy}) \).
Rewrite \( \sqrt{xy} \) as \( (xy)^{1/2} \):
\( \log \left(\frac{x - y}{7}\right) = \log (xy)^{1/2} \).
Apply the power rule on the right side:
\( \log \left(\frac{x - y}{7}\right) = \frac{1}{2} \log (xy) \).
Apply the product rule on the right side:
\( \log \left(\frac{x - y}{7}\right) = \frac{1}{2} (\log x + \log y) \).
This proves the required identity. Manipulating the algebraic expression to match the logarithmic form is key.
In simple words: Start with \( x^2 + y^2 = 51xy \). Subtract \( 2xy \) from both sides to get \( (x-y)^2 = 49xy \). Divide by \( 49 \) and take the square root to get \( \frac{x-y}{7} = \sqrt{xy} \). Then, take the \( \log \) of both sides and use log rules to show it matches \( \frac{1}{2} (\log x + \log y) \).

🎯 Exam Tip: This type of question requires strong algebraic manipulation skills before applying logarithmic properties. Remember to choose the positive square root when dealing with arguments of logarithms.

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